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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Solution:</dfn> The characteristic equation is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
r^4-1=0 \to (r^2+1)(r^2-1)=0 \to (r-1)(r+1)(r-i)(r+i)=0.
\end{equation*}
</div>
<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
r_1=1,\quad r_2=-1,\quad r_3=i,\quad r_4=-i.
\end{equation*}
</div>
<p class="continuation">There are four solutions:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
e^x, \quad e^{-x},\quad e^{i x},\quad e^{-ix}.
\end{equation*}
</div>
<p class="continuation">The last two can be replaced by <span class="process-math">\(e^{\lambda x} \cos \mu x=\cos x\)</span> and <span class="process-math">\(e^{\lambda x} \sin \mu x=\sin x\text{.}\)</span> Therefore, the general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=C_1 e^x+C_2 e^{-x}+C_3 \cos x+C_4 \sin x.
\end{equation*}
</div>
<p class="continuation">From the initial conditions, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp; y(0)=1 \to C_1+C_2+C_3=1,\\
&amp;y^{\prime}(0)=0 \to C_1-C_2+C_4=0,\\
&amp;y^{\prime \prime}(0)=-1 \to C_1+C_2-C_3=-1,\\
&amp;y^{\prime \prime \prime}(0)=0 \to C_1-C_2-C_4
=0.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">This gives</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
C_1=C_2=C_4=0,\quad C_3=1.
\end{equation*}
</div>
<p class="continuation">Thus, the solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=\cos x.
\end{equation*}
</div>
<span class="incontext"><a href="sec4_2.html#p-163" class="internal">in-context</a></span>
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